Show that $\{x \in \mathbb{Q}:x \geq 0, x^2 \leq 2\}$ has no rational least upper bound. Lets denote the least upper bound by  $\alpha \in \mathbb{Q}$ and $\delta > 0$ be a small number.
Now $\alpha^2 \neq 2$ because there is no such rational $\alpha$. If $\alpha^2 > 2$ then $(\alpha +\delta)^2 >2$ and so $\alpha$ is not a least upper bound. 
I can't obtain a valid reason why $\alpha^2 < 2$ can not be the case. 
 A: At first i can't follow your argument for why if $\alpha^2>2$ than $\alpha$ can't be a least upper bound, because you only give another upper bound which is greater. 
If $\alpha^2 <2$ then you have to find a rational number $q$ such that $\alpha^2< q^2< 2$.
First we show that if $\alpha^2>2$ then $\alpha$ can't be a least upper bound. 
Because of $\alpha^2 >2$ there is a $\varepsilon >0$ such that $\alpha^2 > 2+\varepsilon$. Due to archimedes wlog we may assume this $\varepsilon$ be be rational. We now want to find a $\delta >0$ such that $\alpha-\delta$ is another upper bound, which would contradict that $\alpha$ is the least upper bound. We now have 
\begin{align*}
(\alpha -\delta)^2 &= \alpha^2 - \delta(2 \alpha -\delta)\\
 & > 2 + \varepsilon - \delta (2 \alpha - \delta)
\end{align*}
which tells us we want to have that 
$$\varepsilon > \delta(2\alpha-\delta)$$ 
to show that for this choice of $\delta$ the number $\alpha-\delta$ will be an upper bound for our set. This calculation even shows, that not every $\varepsilon$ should be neglected. Even though you could solve this inequality exactly due to the easy terms, it is even more easy (and should be trained) to make some more estimates.
As $\delta >0$ (and $\alpha >0$ we choose $\delta$ such that
$$\varepsilon  > 2\delta \alpha$$ in other words
$\delta < \frac{\varepsilon}{2 \alpha}$ we see that
$$\varepsilon > 2\delta \alpha > \delta( 2\alpha -\delta)$$
and hence
 $\alpha-\delta$ is a lower upper bound.
